What is the integral from 0 to 1/2 of 1/(1-x^2) ?

The integral from 0 to 1/2 of 1/(1-x^2) is (ln(3))/2

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integral from 0 to 1/2 of 1/(1-x^2)

\int_{0}^{\frac{1}{2}} \frac{1}{1 - x ^{2}} d x

\frac{ln ( 3 )}{2}

Decimal Notation

0.54930 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{1}{1 - x ^{2}} d x

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = \frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}

= [\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{0}^{\frac{1}{2}}

Simplify

[\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}]_{0}^{\frac{1}{2}} : [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{\frac{1}{2}}

= [\frac{1}{2} (ln ∣ x + 1 ∣ - ln ∣ x - 1 ∣)]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

= \frac{ln ( 2 ) + ln ( \frac{3}{2} )}{2}

Simplify

= \frac{ln ( 3 )}{2}

\int_{0}^{\frac{π}{2}} tan^{2} (θ) d θ

\int_{1}^{4} 5 d x

\int_{1}^{4} 2 d x

\int_{0}^{\frac{1}{2}} x d x

\int_{1}^{4} 2 x + 1 d x

What is the integral from 0 to 1/2 of 1/(1-x^2) ?
The integral from 0 to 1/2 of 1/(1-x^2) is (ln(3))/2